Consider a Taylor-expansion of \(\tilde{U}_n (\theta )\)  centered at \(\theta _0\)  when \(\hat{\theta }_n \in \Theta _0\) . We can write as
\(\tilde{U}_n (\hat{\theta }_n) &= \tilde{U}_n (\theta _0) - (\hat{\theta }_n - \theta _0) \frac{\partial }{\partial \theta } \tilde{U}_n (\theta ) \mid _{\theta =\theta _0} - (\hat{\theta }_n - \theta _0) \sum _{l=1}^p (\hat{\theta }_{ln} - \theta _{l0}) \frac{\partial ^2}{\partial \theta \partial \theta _l} \tilde{U}_n (\theta ) \mid _{\theta =\theta ^\star } \\0 &= \tilde{U}_n (\theta _0) + (\hat{\theta }_n - \theta _0) \left[ \frac{\partial }{\partial \theta } \tilde{U}_n (\theta ) \mid _{\theta =\theta _0} + \frac{1}{2} \sum _{l=1}^p (\hat{\theta }_{ln} - \theta _{l0}) \frac{\partial ^2}{\partial \theta \partial \theta _l} \tilde{U}_n (\theta ) \mid _{\theta =\theta ^\star } \right] \\n^{1/2} (\hat{\theta }_n - \theta _0) &= \left[ \frac{1}{n} \frac{\partial }{\partial \theta } \tilde{U}_n (\theta ) \mid _{\theta =\theta _0} + \frac{1}{2n} \sum _{l=1}^p (\hat{\theta }_{ln} - \theta _{l0}) \frac{\partial ^2}{\partial \theta \partial \theta _l} \tilde{U}_n (\theta ) \mid _{\theta =\theta ^\star } \right]^{-1} n^{-1/2} \tilde{U}_n (\theta _0)\) 

where \(\theta ^\star \)  is on the line segment between \(\hat{\theta }_n\)  and \(\theta _0\) . First, the term
\(\frac{1}{n} \frac{\partial }{\partial \theta } \tilde{U}_n (\theta ) \mid _{\theta =\theta _0} + \frac{1}{2n} \sum _{l=1}^p (\hat{\theta }_{ln} - \theta _{l0}) \frac{\partial ^2}{\partial \theta \partial \theta _l} \tilde{U}_n (\theta ) \mid _{\theta =\theta ^\star }\overset{P}{\rightarrow } \Xi (\theta _0)\) 
under conditions VI.1.1(A)–(E) in [1]} and Assumptions REF –REF . To see this, note the first term converges to \(\Xi (\theta _0)\)  while the second term is bounded in probability by \(p M \Vert  \theta ^\star - \theta _0 \Vert \)  for some finite constant \(M\)  independent of \(\theta ^\star \)  where \(\Vert  \cdot \Vert \)  is the Euclidean norm. The second term therefore goes to zero since \(\Vert  \theta ^\star - \theta _0 \Vert  \rightarrow 0\)  as \(n \rightarrow \infty \) .
We now consider \(n^{-1/2} \tilde{U}_n (\theta _0 )\) . Recall \(\tilde{U}_n (\theta _0)\)  can be decomposed into three terms as in (REF ). Here, we assume \(K_x \rightarrow \infty \)  so the final term is negligible. This leaves two terms
\(U_n (\theta _0) = \sum _{i=1}^n \int _0^\tau \frac{h^{(1)}(u; \theta )}{h (u;\theta )} dM_i^{(e)} (t)\) 
where \(N_i^{(e)} (t)\)  and \(N_i^{(s)} (t)\)  are the orthogonal event and subsampling counting processes with jumps at times \(t \in {\bf T}_i\)  and \(t \in {\bf D}_i\)  respectively.
and
\((\hat{U}_n (\theta _0) - U_n (\theta _0) )&= \sum _{i=1}^n \int _0^\tau \frac{h^{(1)}(s; \theta _0)}{\pi _i (s) + h(s; \theta _0)} dM_i (s) \\&= \sum _{i=1}^n \int _0^\tau \frac{h^{(1)}(s; \theta _0)}{\pi _i (s) + h(s; \theta _0)} dM^{(e)}_i (s) +\sum _{i=1}^n \int _0^\tau \frac{h^{(1)}(s; \theta _0)}{\pi _i (s) + h(s; \theta _0)} dM_i^{(s)} (s) \\&= \sum _{i=1}^n \int _0^\tau (1-w(s;\theta _0)) \cdot \frac{h^{(1)}(s; \theta _0)}{h_i (s;\theta _0)} dM_i^{(e)} (s) +\sum _{i=1}^n \int _0^\tau w(s;\theta _0) \cdot \frac{h^{(1)}(s; \theta _0)}{\pi _i (s)} dM_i^{(s)} (s)\) 

Taking difference with \(U_n(\theta _0)\) , the first term becomes
\(\sum _{i=1}^n \int _0^\tau w(s;\theta _0) \cdot \frac{h^{(1)}(s; \theta _0)}{h_i (s;\theta _0)} dM_i^{(e)} (s)\) 
These are two orthogonal square integrable martingales with quadratic variation given by
\(\sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \left[ \frac{h^{(1)}(s; \theta _0)}{h_i (s;\theta _0)} \right] \times \left[ \frac{h^{(1)} (s;\theta _0)}{h_i (s;\theta _0)} \right]^\top h_i (s) ds\) 
and
\(\sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \left[ \frac{h^{(1)}(s; \theta _0)}{\pi _i (s)} \right] \times \left[ \frac{h^{(1)} (s;\theta _0)}{\pi _i (s)} \right]^\top \pi _i (s) ds\) 
respectively. Using the definition \(w(s;\theta _0) = \pi _i(s) / (\pi _i(s)+h_i(s;\theta _0))\) , we have
\(&\frac{1}{n} \sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \left[ \frac{h^{(1)}(s; \theta _0)}{h_i (s;\theta _0)} \right] \times \left[ \frac{h^{(1)} (s;\theta _0)}{h_i (s;\theta _0)} \right]^\top h_i (s) ds \\&+\frac{1}{n} \sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \left[ \frac{h^{(1)}(s; \theta _0)}{\pi _i (s)} \right] \times \left[ \frac{h^{(1)} (s;\theta _0)}{\pi _i (s)} \right]^\top \pi _i (s) ds \\=& \frac{1}{n} \sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \frac{h^{(1)}(s; \theta _0) [h^{(1)} (s;\theta _0)]^\top }{h_i (s;\theta _0)} \left[ 1 + \frac{h(s;\theta _0)}{\pi _i (s)} \right] ds \\=& \frac{1}{n} \sum _{i=1}^n \int _0^\tau w^2(s;\theta _0) \frac{h^{(1)}(s; \theta _0) [h^{(1)} (s;\theta _0)]^\top }{h_i (s;\theta _0)} \left[ \frac{\pi _i (s) + h(s;\theta _0)}{\pi _i (s)} \right] ds \\=& \frac{1}{n} \sum _{i=1}^n \int _0^\tau w(s;\theta _0) \frac{h^{(1)}(s; \theta _0) [h^{(1)} (s;\theta _0)]^\top }{h_i (s;\theta _0)} ds\rightarrow \Xi (\theta _0).\) 

Returning to the Taylor-expansion, under conditions VI.1.1(A)–(E) in [1]} and Assumptions REF –REF , we can apply Rebolledo's martingale central limit theorem so that
\(\sqrt{n} (\hat{\theta }_n - \theta _0) \overset{D}{\rightarrow } N(0, \Xi (\theta _0) ^{-1} \times \Xi (\theta _0) \times \Xi (\theta _0) ^{-1} )\) 
which completes the proof.
